Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer:
The molecular formula of cacodyl is C₄H₁₂As₂.
Explanation:
<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:
- 209.96 g * 22.88/100 = 48.04 g C
- 209.96 g * 5.76/100 = 12.09 g H
- 209.96 g * 71.36/100 = 149.83 g As
Now we convert those masses into moles:
- 48.04 g C ÷ 12 g/mol = 4.00 mol C
- 12.09 g H ÷ 1 g/mol = 12.09 mol H
- 149.83 g As ÷ 74.92 g/mol = 2.00 mol As
Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.
let me know when u find out plz because i would like to know as well its one of my chemistry qustions in an assiment. :)
Answer: (3) 15
Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.
From the molarity and volume of HClO4, we can determine how many moles of H+ we initially have:
0.18 M HClO4 * 0.100 L HClO4 = 0.018 moles H+
We can determine how many moles of OH- we have from the molarity and volume of LiOH:
0.27 M LiOH * 0.030 L LiOH = 0.0081 moles OH-
When the HClO4 and LiOH neutralize each other, the remaining will be
0.018 moles H+ - 0.0081 moles OH- = 0.0099 moles of excess H+
This means that the molarity [H+] will be
[H+] = 0.0099 moles H+ / (0.100 L + 0.030 L) = 0.07615 M
The pH of the solution will therefore be
pH = -log [H+] = -log 0.07615 = 1.12
